Input-output Controllability Analysis

Transcription

1 Input-output Controllability Analysis Idea: Find out how well the process can be controlled - without having to design a specific controller Note: Some processes are impossible to control Reference: S. Skogestad, ``A procedure for SISO controllability analysis - with application to design of ph neutralization processes'', Comp.Chem.Engng., 20, , WANT TO QUANTIFY! 1

2 Rules Rules 1-3: speed of response Rule 1: Fast response required to reject large disturbance BUT (rule 2): Response time is limited by effective time delay Rule 3: Fast response needed for stabilization Rule 4: Input constraints Large disturbances may give input saturation Rules for speed of response (assuming control with integral action) Define ω c =1/τ c = closed-loop bandwidth = where L is approx. 1 Define ω d as frequency where g d =1 (scaled model, frequency where y =1 for d =1) Rule 1: Fast response required to reject large disturbance Need ω c >ω d (τ c < 1/ω d ) Rule 1 is for typical case where g d is highest at low frequencies The more exact rule is: We need Sg d <1, or approximately: L > g d at frequencies where g d >1. Rule 2: Response time is limited by effective time delay Need ω c < 1/θ (τ c > θ. SIMC-rule!) Where θ is effective time delay Rule 3: Fast response needed for stabilization Need ω c > p (τ c < 1/p) Where p is unstable pole, g(s) =k/(s-p) Rule 4: Input constraints: Large disturbances may give input saturation With scaled model: Need G > G d at frequencies where G d >1 This situation is OK according to rules 1-3: p ω d 1/ θ ω c must be in this range ω 2

3 Comment: Ideal controller inverts the plant y = g(s)u + d Ideal controller inverts the plant g(s): Think feedforward, u = c ff (s) (y s -d) Perfect control: want y=y s c ff = 1/g(s) = g -1 Limitations on perfect control: Inverse cannot always be realized: 1. Input saturation, u > u max 2. Time delay, g=e -θs. g -1 = e θs = prediction (not possible) Solution: Omit 3. Inverse response, g = -Ts+1. g -1 = 1/(-Ts+1) = unstable (not possible as u will be unbounded) Solution: Omit 4. More poles than zeros, g = 1/(τs+1), g-1 = τs + 1 = pure differentiation (not possible as u will be unbounded). Solution: Replace by: (τs + 1)/(τ c +1) where τ c < τ is a tuning parameter Example. g(s) = 5 (-0.5s+1) e-2s / (3s+1). Realizable inverse (feedforward): 0.2 (3s+1)/(τ c +1). E.g. choose τ c =0.5 So we know what limits us from having perfect control Same limitations apply to feedback control Controllability analysis: Want to find out what these limitations imply in terms of acceptable control, y-y s < y max We use scaled model Rules 1 and 4 3

4 Recall: Closed-loop frequency response (S) Rule 1 e 10 1 SIMC: M s =1.70 ZN: M s = 2.93 S SIMC ZN Control: GOOD BAD NO EFFECT! w = logspace(-2,1,1000); [mag1,phase]=bode(1/(1+l1),w); [mag2,phase]=bode(1/(1+l2),w); figure(1), loglog(w,mag1(:),'red',w,mag2(:),'blue',w,1,'-.') axis([0.01,10,0.001,10]) MAIN REASON FOR CONTROL: DISTURBANCES! Rule 1 Disturbances and Loop gain L S = 1/(1+L) where L = gc No control («open-loop»): y = g u + g d d With control: y = S g d d Scaled variables: Want Sg d <1 at all ω Approximation at low frequencies where L is large: S = 1/L So want (in scaled variables): L > g d Up to about frequency ω c where L =1 4

5 Time domain Rule 1 Consider response to step disturbance, g d = k d /(τ d s+1) Output reaches Δy = (k d θ /τ d ) Δd at time θ (approximately) If this is larger than acceptable then we are in trouble If Δd=1 and requirement is Δy < 1 then we must require k d /τ d < 1/θ (combined rule 1+2) ω d } Easier to generalize in frequency domain Consider disturbance d(t)=sinωt MAIN REASON FOR CONTROL: DISTURBANCES! Rule 4 Need control up to frequency w d where G d =1 -> Need w c > w d (w c is frequency where L =1) L=gc G d ωc ω d Note: Have ω c =1/τ c 5

6 Rule 4 Rule 4: Rules for speed of response (assuming control with integral action) Define ω c =1/τ c = closed-loop bandwidth = where L is approx. 1 Define ω d as frequency where g d =1 (scaled model, frequency where y =1 for d =1) Rule 1: Fast response required to reject large disturbance Need ω c >ω d (τ c < 1/ω d ) Rule 1 is for typical case where g d is highest at low frequencies The more exact rule is: We need Sg d <1, or approximately: L > g d at frequencies where g d >1. Rule 2: Response time is limited by effective time delay Need ω c < 1/θ (τ c > θ. SIMC-rule!) Where θ is effective time delay Rule 3: Fast response needed for stabilization Need ω c > p (τ c < 1/p) Where p is unstable pole, g(s) =k/(s-p) Rule 4: Input constraints: Large disturbances may give input saturation With scaled model: Need G > G d at frequencies where G d >1 This situation is OK according to rules 1-3: p ω d 1/ θ ω c must be in this range ω 6

7 10 3 EXAMPLE, g = 500/((50*s+1)*(10*s+1)) G 10 2 gd = 9/(10*s+1) 10 1 G d ω d = s=tf('s') g = 500/((50*s+1)*(10*s+1)) gd = 9/(10*s+1) w = logspace(-3,1,1000); [mag,phase]=bode(g,w); [magd,phased]=bode(gd,w); loglog(w,mag(:),'blue',w,magd(:),'red',w,1,'black'), grid on Effective delay θ limits ω c PI-control: ω c <1/θ = 1/5 = 0.2 PID-control: ω c <1/θ = 1/0 = CHECK CONTROLLABILITY ANALYSIS WITH SIMULATIONS 7

8 y PI control not acceptable* s=tf('s') g = 500/((50*s+1)*(10*s+1)) gd = 9/(10*s+1) % SIMC-PI with tauc=theta=5 Kc=(1/500)*(55/(5+5)); taui=55; taud=0; ± *As expected since need ω c > ω d = 0.9, but can only achieve ω c <1/θ = 1/5 = 0.2 PID control acceptable: y and u are within ± y u g = 500/((50*s+1)*(10*s+1)) gd = 9/(10*s+1) %SIMC-PID (cascade form) with tauc=1/wd=1: Kc=(1/500)*(50/(1+0)); taui=50; taud=10; 8

9 If process is not controllable: Need to change the design For example, dampen disturbance by adding buffer tank: Level control unimportant, but need good mixing Level control is NOT tight -> level varies Integral action is not recommended for averaging level control Problem 1 9

10 Problem 2 Problem 3-10

11 Problem 4 g = 200/((20*s+1)*(10*s+1)*(s+1)) gd = 4/((3*s+1)*(s+1)^3) Kc=(1/200)*20/1,taui=20,taud=10.5 Problem 5-11

12 Problem 6 ph-neutralization. y = c H+ - c OH- (want= mol/l,ph=7+-1) u = q base (c OH- =10mol/l, ph=15) d = q acid (c H+ =10mol/l, ph=-1) Using tanks in series, Acid and base in tank 1. G d Scaled model: k d = 2.5e6 Each tank: τ = 1000s Control: θ = 10s (meas. delay for ph) Problem: How many tanks? n=1 n=2 n=3 ω [rad/s] Reference for more applications of controllability analysis: Chapter 5 in book by Skogestad and Postlethwaite (2005) Control system 3 tanks: Neutralization (base addition) only in tank 1 gives large effective delay (>> 10s) because of tank dynamics in g(s) Suggested solution is to add (a little) base also in the other tanks: ph 2 ph 5 12

13 Conclusion Use controllability analysis To avoid spending time on impossible control problem To help design the process (e.g., size buffer tanks) Also useful for tuning. τ c = SIMC tuning parameter = 1/ω c Must for acceptable controllability have: Agrees with SIMC-rules Tight control: τ c = θ Smooth control: τ c = 1/ω d Exam Wednesday 06 December 2017 from 09:00 to 13:00 The test (questions) is in English but you may answer in Norwegian or English. Permitted examination support material: One (1) A4 double-sided piece of paper with your handwritten notes (it does not need to be approved or stamped prior to the exam). No other written material. Standard calculator. Note: Remember to state clearly all assumptions you make. Q&A session Alternative: Monday 04 Dec